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lgbasallote

Prove that if n is a perfect square, then n +2 is not a perfect square

  • one year ago
  • one year ago

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  1. lgbasallote
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    i suppose first step would be to assume n is perfect square

    • one year ago
  2. lgbasallote
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    so n = m^2

    • one year ago
  3. lgbasallote
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    then i suppose i find n^2

    • one year ago
  4. lgbasallote
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    i mean n + 2

    • one year ago
  5. lgbasallote
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    n + 2 would be m^2 + 2

    • one year ago
  6. lgbasallote
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    so now... i assume m^2 + 2 is a perfect square...

    • one year ago
  7. mukushla
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    we have no 2 perfect square numbers with difference 2

    • one year ago
  8. lgbasallote
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    so i let m^2 + 2 = k^2

    • one year ago
  9. lgbasallote
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    @mukushla that's not really a proof.....

    • one year ago
  10. shubhamsrg
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    you may continue @igba

    • one year ago
  11. lgbasallote
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    hmm i suppose the next step would be m^2 - k^2 = 2 so (m+k)(m-k) = 2 i don't think this proves anything...

    • one year ago
  12. mukushla
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    actually that is :) \[m^2-k^2=2\]\[(m-k)(m+k)=2\]and this is impossible

    • one year ago
  13. lgbasallote
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    why so?

    • one year ago
  14. shubhamsrg
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    well,,you've almost proved it.. now m and k both are integers/.

    • one year ago
  15. shubhamsrg
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    and its k^2 -m^2 = 2 and not otherwise..

    • one year ago
  16. lgbasallote
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    oh yes

    • one year ago
  17. lgbasallote
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    2 = (k + m)(k-m)

    • one year ago
  18. lgbasallote
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    now...integers huh...

    • one year ago
  19. mukushla
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    \[m+k>m-k\]so\[m+k=2\]so\[m=k=1\]but it gives\[m-k=0\]

    • one year ago
  20. shubhamsrg
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    you know k>m.. so from (k+m)(k-m)= 1*2 =>k+m =2 and k-m = 1 this doesnt have an integral solution..

    • one year ago
  21. lgbasallote
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    why do i know k > m again?

    • one year ago
  22. shubhamsrg
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    m^2 + 2 = k^2 both m and k are positive integers hence k>m

    • one year ago
  23. shubhamsrg
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    in layman language,, k^2 - m^2 >0

    • one year ago
  24. lgbasallote
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    when was it assumed that m and k are positive?

    • one year ago
  25. shubhamsrg
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    what else do you mean by perfect square ?

    • one year ago
  26. lgbasallote
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    i don't see how that relates.... the definition of a perfect square is x = k^2

    • one year ago
  27. shubhamsrg
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    4 is a perfect sq since its 2^2 i.e. square of a positive integer.. and likewise..

    • one year ago
  28. estudier
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    @mukushla "we have no 2 perfect square numbers with difference 2" Because 0,1,4,9,6,5,6,9,4,1 etc....

    • one year ago
  29. lgbasallote
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    (-2)^2 is also 4...

    • one year ago
  30. shubhamsrg
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    you take the absolute value

    • one year ago
  31. shubhamsrg
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    i mean you have to..

    • one year ago
  32. lgbasallote
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    why?

    • one year ago
  33. shubhamsrg
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    see,, (k+m)(k-m) =2 either both are positive are both are negative,,since RHS is postive.. so even if you take -ve values,, negative * negative = positive ,,i.e. treated as absolute value..

    • one year ago
  34. lgbasallote
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    ...i don't think this is a valid proof....

    • one year ago
  35. shubhamsrg
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    another explanation might me x is a perfect square because its y^2 x = y^2 = (-y)^2 sqrt(x) = | y | = | -y |

    • one year ago
  36. jhonyy9
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    who can writing here now two perfect squares with difference of 2 ? so because i think do not exist or .... ???

    • one year ago
  37. lgbasallote
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    since when did sqrt x become |-y|

    • one year ago
  38. estudier
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    0 to 9 squares end in 0,1,4,9,6,5,6,9,4,1 and pattern repeats. No +2 in there.

    • one year ago
  39. jhonyy9
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    so and i think that this difference of two will be the ,,key" of this proof of ...

    • one year ago
  40. jhonyy9
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    @lgbasallote what is your opinion from this ,please ?

    • one year ago
  41. lgbasallote
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    my opinion is that this is some tricky algebra

    • one year ago
  42. jhonyy9
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    how do you think it ???

    • one year ago
  43. jhonyy9
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    so and why ?

    • one year ago
  44. lgbasallote
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    because i can't think of a proof

    • one year ago
  45. lgbasallote
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    how can you prove/disprove that the difference of two squares is 2...

    • one year ago
  46. lgbasallote
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    hmmm...wait...

    • one year ago
  47. jhonyy9
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    so i think this is very easy logicaly but to prove it will be difficile i think so but there are again @saifoo.khan and @satellite73 probably they will can doing it hope so much

    • one year ago
  48. lgbasallote
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    their fields of specialty aren't exactly in proving

    • one year ago
  49. estudier
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    What's wrong with my proof?

    • one year ago
  50. lgbasallote
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    anyway... m^2 + 2 = k^2 k^2 - m^2 = 2 for it to be equal to 2...k^2 and m^2 should either be both even...or both odd so if i assume both are even.. then k and m are also both even so i can rewrite this as (2x)^2 - (2y^2) = 2 4x^2 - 4y^2 = 2 then... 2x^2 - 2y^2 = 1 x^2 - y^2 = 1/2 then since it's x^2 and y^2 are both integers...the difference should also be integer \('\therefore\) contradiction

    • one year ago
  51. lgbasallote
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    and i believe your proof @jhonyy9 did not involve proving...but brute force substitution..

    • one year ago
  52. lgbasallote
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    @estudier i mean

    • one year ago
  53. lgbasallote
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    @mukushla too

    • one year ago
  54. estudier
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    So, that doesn't mean it's invalid..

    • one year ago
  55. lgbasallote
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    it does actually.....you can't do infinite substitutions to prove anything....

    • one year ago
  56. estudier
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    There is no infinity involved, all integer squares end in the numbers I gave.

    • one year ago
  57. lgbasallote
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    infinite loops involves infinity agree?

    • one year ago
  58. estudier
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    No infinite loop either, just logic.

    • one year ago
  59. estudier
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    An integer ends in one of digits 0 to 9. The squares must end in etc....

    • one year ago
  60. lgbasallote
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    but we both know that setting values is illegal in proving

    • one year ago
  61. estudier
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    I am not setting any values.

    • one year ago
  62. lgbasallote
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    and actually... you were checking 1^2, 2^2, 3^2, 4^2, etc. thus that will go on forever

    • one year ago
  63. lgbasallote
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    you're checking the squares of each number

    • one year ago
  64. estudier
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    I am not.

    • one year ago
  65. lgbasallote
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    and though there is a pattern...there is still susbtitution involved

    • one year ago
  66. lgbasallote
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    and then you'll have to substitute forever to verify that the pattern really does not break

    • one year ago
  67. lgbasallote
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    then infinity is involved

    • one year ago
  68. estudier
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    It's not a pattern, it's a fact 0 to 9 squares end in 0,1,4,9,6,5,6,9,4,1

    • one year ago
  69. lgbasallote
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    so you do admit that you checked the 0 to 9 squares

    • one year ago
  70. estudier
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    Anything else is impossible.

    • one year ago
  71. lgbasallote
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    ...that's substitution

    • one year ago
  72. lgbasallote
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    anyway...my point is...substitution isn't really allowed in proving

    • one year ago
  73. estudier
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    You will find that this fact is used a lot in many valid proofs in number theory.

    • one year ago
  74. lgbasallote
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    perhaps

    • one year ago
  75. estudier
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    OK, I can also do the algebra proof, but that doesn't mean it's any better than mine.

    • one year ago
  76. lgbasallote
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    since algebra is applicable in all numbers set to it....wouldn't that make it better than brute force?

    • one year ago
  77. estudier
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    I am not using brute force, brute force is something like testing or experimental math. Or exhaustion (which is also a valid proof technique). It's the use of a fact about squares (ALL squares)

    • one year ago
  78. lgbasallote
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    can you prove it? that it's a fact about squares?

    • one year ago
  79. estudier
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    Yes, all numbers end in 0 to 9 and so their squares end in .....QED.

    • one year ago
  80. lgbasallote
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    you didn't really demonstrate anything...you just repeated what you said

    • one year ago
  81. estudier
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    You asked for proof, I just gave u a proof.

    • one year ago
  82. estudier
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    If you deny my proof, provide a counterexample.

    • one year ago
  83. lgbasallote
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    since you try so hard to defend your proof, i'll accept it

    • one year ago
  84. estudier
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    That's very gracious, thank you:-)

    • one year ago
  85. lgbasallote
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    do check on my algebra proof though please. i would like to know if that's the way contradiction works

    • one year ago
  86. estudier
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    OK, I will look at it now....

    • one year ago
  87. estudier
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    Did u do the case when both are odd? (Perhaps a more straightforward way is one you start with n= some k squared then next up must be (k+1)^2 = k^2 + 2k +1 and see what you can deduce from that...)

    • one year ago
  88. lgbasallote
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    why do i have to do odd too? if i already disproved using even...doesn't that prove by contradiction already?

    • one year ago
  89. estudier
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    No because you divided the problem in cases, means you have to prove both (else you are missing all those cases)

    • one year ago
  90. lgbasallote
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    oh i see. but that kind of proof is right? because i'm a little skeptic if stating "2 can only be an outcome of even and even or odd and odd" is a valid statement

    • one year ago
  91. lgbasallote
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    again i ask... how is k > m

    • one year ago
  92. estudier
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    (k+1)^2 = k^2 + 2k +1 >= n +3 > n+2 QED

    • one year ago
  93. mukushla
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    factoring involves lesser steps...its obvious that k>m \[k^2−m^2=2\]\[(k−m)(k+m)=2\]easily u can see \[k-m=1\]\[k+m=2\]add them\[2k=3\] Contradiction

    • one year ago
  94. lgbasallote
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    again i ask for the nth time @mukushla how is k > m

    • one year ago
  95. mukushla
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    ok man i just want to answer ur question

    • one year ago
  96. estudier
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    "2 can only be an outcome of even and even or odd and odd" is a valid statement Why do you think it might not be valid?

    • one year ago
  97. lgbasallote
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    well because 8 - 4 is not 2 but it's even minus even

    • one year ago
  98. lgbasallote
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    @mukushla i really don't get why k > m... i know k^2 > m^2 but does that make k > m?

    • one year ago
  99. estudier
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    OK, so finally, the algebra Given n=k^2 -> (k+1)^2 = k^2 + 2k +1 >= n +3 > n+2 QED

    • one year ago
  100. mukushla
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    i thought they are positive :) sorry

    • one year ago
  101. lgbasallote
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    why (k+1)^2 @estudier ?

    • one year ago
  102. estudier
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    I already, said, that's the next square....

    • one year ago
  103. lgbasallote
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    next square?

    • one year ago
  104. lgbasallote
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    is this still related to my proof?

    • one year ago
  105. estudier
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    ? n is k^2, what is the next square?

    • one year ago
  106. mukushla
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    no problem ... this is better\[|k|^2-|m|^2=2\]\[(|k|-|m|)(|k|+|m|)=2\]so\[|k|-|m|=1\]\[|k|+|m|=2\]because this time im sure that \(|k|>|m|\) ...now add them\[2|k|=3\]Contradiction and we are done

    • one year ago
  107. estudier
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    No, this proof Given n=k^2 -> (k+1)^2 = k^2 + 2k +1 >= n +3 > n+2 QED

    • one year ago
  108. lgbasallote
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    are you allowed to do that? @mukushla

    • one year ago
  109. lgbasallote
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    can you write it in a vertical fashion @estudier ? the signs are confusing me where the steps end

    • one year ago
  110. mukushla
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    why not...for any real number \[a^2=|a|^2\]

    • one year ago
  111. estudier
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    Given n=k^2 -> (k+1)^2 = k^2 + 2k +1 k^2 + 2k +1 >= n +3 n +3 > n+2 QED

    • one year ago
  112. estudier
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    That's 2 proofs I have given now...

    • one year ago
  113. lgbasallote
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    a few questions in mind @estudier 1) why find the next square? 2) how is k^2 + 2k + 1 > = n + 3 3) what does n + 3 > n+2 prove?

    • one year ago
  114. estudier
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    Think about it....:-)

    • one year ago
  115. lgbasallote
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    in the meanwhile... @mukushla why |k| - |m| = 1?

    • one year ago
  116. mukushla
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    because\[|k|-|m|<|k|+|m|\]

    • one year ago
  117. lgbasallote
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    how does that make |k| - |m| = 1

    • one year ago
  118. mukushla
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    u have 2 positive integers with ab=2 and a<b what are a and b....clearly 1 and 2

    • one year ago
  119. mukushla
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    right?

    • one year ago
  120. lgbasallote
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    who said those are positive?

    • one year ago
  121. mukushla
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    |k|>|m| right?

    • one year ago
  122. lgbasallote
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    something feels wrong...

    • one year ago
  123. lgbasallote
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    anyway...where did you get 2|k| > 3

    • one year ago
  124. mukushla
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    ok...think on what i said...because i believe thats the shortest way to prove ur statement

    • one year ago
  125. lgbasallote
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    |2k| = 3 *

    • one year ago
  126. mukushla
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    adding\[|k|-|m|=1\]and\[|k|+|m|=2\]

    • one year ago
  127. estudier
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    Going back to your question about 2, u can just adjust it to difference of 2q...

    • one year ago
  128. lgbasallote
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    im really skeptical about this... especially that |k|^2 - |m|^2 = 2

    • one year ago
  129. lgbasallote
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    difference of 2q?

    • one year ago
  130. estudier
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    You were worried about the validity of your statement, change it to 2q instead of just 2.

    • one year ago
  131. estudier
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    Or 2p, or whatever letter u like best.

    • one year ago
  132. estudier
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    1) why find the next square? Because we want to see that it isn't n+2 2) how is k^2 + 2k + 1 > = n + 3 Substitute n=k^2 3) what does n + 3 > n+2 prove? See 1)

    • one year ago
  133. lgbasallote
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    oh. when i said difference of two numbers equal to 2

    • one year ago
  134. lgbasallote
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    but wait...if n + 2 were two squares away....wouldn't that make finding the next square invalid?

    • one year ago
  135. lgbasallote
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    and @mukushla how did you get |k| > |m|

    • one year ago
  136. estudier
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    No, because the next square is (k+1)^2 (which might or might not be equal to n+2, that is what we have to discover.

    • one year ago
  137. lgbasallote
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    and since the next square is greater than n + 2...that makes n + 2 invalid for any succeeding squares?

    • one year ago
  138. estudier
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    Yes, the algebra is completely general...

    • one year ago
  139. lgbasallote
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    i'm still confused about k^2 + 2k + 1 though...if n = k^2...it just makes n + 2k + 1...

    • one year ago
  140. estudier
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    ...which is greater than n+3, right?

    • one year ago
  141. lgbasallote
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    it is?

    • one year ago
  142. estudier
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    Number theorists don't pay attention to 0

    • one year ago
  143. lgbasallote
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    unless k is 0....

    • one year ago
  144. lgbasallote
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    why not?

    • one year ago
  145. estudier
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    Tradition, convention, call it as u like..... So in fact there are 2 definitions for natural number , one with 0 and one without.

    • one year ago
  146. sauravshakya
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    Can I try?

    • one year ago
  147. lgbasallote
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    ..you give so many weird proofs......the sad part is that i cannot disprove it because i know nothing about it....

    • one year ago
  148. estudier
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    More the merrier.

    • one year ago
  149. estudier
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    "because i know nothing about it" But you do now:-)

    • one year ago
  150. lgbasallote
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    it's not like it's limited to the 99s....

    • one year ago
  151. lgbasallote
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    i was referring to your claims earlier like brute force is valid proof in number theory and that number theorists don't consider 0...i cannot disprove your proofs because i do not know if these claims are true or not

    • one year ago
  152. estudier
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    Number theory is less onerous than real number stuff because it is integer only. But it is actually this difference that causes all the difficulty, students are not accustomed to thinking in a more freewheeling way.

    • one year ago
  153. sauravshakya
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    Let n be a perfect square where n=m^2 then next consecutive perfect square number would be (m+1)^2 Now, (m+1)^2-m^2=2m+1 Since, 2m+1>2 as m is a set of natural numbers. Thus, if n is a perfect square..... n+2 cannot be a perfect square as the difference between two perfect squares is always more than 2.

    • one year ago
  154. estudier
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    Yes, that's similar to the one I gave...

    • one year ago
  155. estudier
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    "number theorists don't consider 0" http://en.wikipedia.org/wiki/Natural_number

    • one year ago
  156. estudier
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    http://mathforum.org/library/drmath/view/63510.html Using 0,1,4, etc...

    • one year ago
  157. jhonyy9
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    ok,so who can answering this question ? ,,the line of perfect squares begin with number ... " what is this number ?

    • one year ago
  158. jhonyy9
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    so and for this your question @lgbasallote i think that for n=m^2 and n+2 for being a perfect square so than n+2=m^2 +2 suppose m^2 +2 =x^2 like a perfect square so than m^2 = x^2 -2 so m1,2 = +/- sqrt(x^2 -2)

    • one year ago
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